Question

The solubility of $A{g}_{2}Cr{O}_4$ is $2\times {10}^{-2}$ mol/litre. Its solubility product is :

• A. $3.2\times {10}^{-5}$
• B. $32\times {10}^{-8}$
• C. $16\times {10}^{-8}$
• D. $3.32\times {10}^{-10}$

Answer 1

The correct option is A $3.2\times {10}^{-5}$

The ionization of silver chromate is as shown below:

$A{g}_{2}Cr{O}_4\rightleftharpoons 2A{g}^{+}+Cr{O}_4^{2-}$

The expression for the solubility product is, $K_{sp}=\left[A{g}^{+}{]}^2\right[Cr{O}_4^{2-}]$.

But $\left[A{g}^{+}\right]=2s$ and $\left[Cr{O}_4^{2-}\right]=s$.

Given, $s=2\times {10}^{-2}M$.

Substituting values in the expression for the solubility product, we get

$K_{sp}=(2s{)}^{2}s=4s^3=4(2\times {10}^{-2}{)}^3=3.2\times {10}^{-5}$.